D - Silver Cow Party

最新推荐文章于 2020-02-02 16:01:48 发布

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本文解决了一个有趣的问题：一群牛从各自的农场出发参加聚会并返回，求所有牛往返所需时间的最大值。采用图论中的最短路径算法进行求解。

D - Silver Cow Party

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: A_i,B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

Sample Output

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

题意：求每个地方的牛到某点聚会再回家的最短距离中距离最大的值。

就是求（各点到某点（E）的最短距离+ 某点（E）到各点的最短距离）中最大距离

思路：最短路变形，，dis[i]=tu[star][i] 改成 dis[i]=tu[i][star]就是第二种情况

判断时改为dis[j]=max(dis[j]，dis[point]+tu[j][point]）即可;

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int INF=99999999;
int tu[1010][1010],book[1010],dis1[1010],dis2[1010];
int main()
{
    int m,n,x,minn,flag,e1,e2,t;
    scanf("%d%d%d",&n,&m,&x);
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            if(i==j) tu[i][j]=0;
            else  tu[i][j]=INF;
    for(int i=0; i<m; i++)
    {
        scanf("%d%d%d",&e1,&e2,&t);
        if(tu[e1][e2]>t)
            tu[e1][e2]=t;
    }
    for(int i=1; i<=n; i++)
    {
        dis1[i]=tu[x][i];
        book[i]=0;
    }
    book[x]=1;
    for(int i=1; i<n; i++)
    {
        minn=INF;
        for(int j=1; j<=n; j++)
        {
            if(dis1[j]<minn&&!book[j])
            {
                minn=dis1[j];
                flag=j;
            }
        }
        book[flag]=1;
        for(int k=1; k<=n; k++)
        {
            dis1[k]=min(dis1[k],dis1[flag]+tu[flag][k]);
        }
    }
    for(int i=1; i<=n; i++)
    {
        dis2[i]=tu[i][x];
        book[i]=0;
    }
    book[x]=1;
    for(int i=1; i<n; i++)
    {
        minn=INF;
        for(int j=1; j<=n; j++)
        {
            if(dis2[j]<minn&&!book[j])
            {
                minn=dis2[j];
                flag=j;
            }
        }
        book[flag]=1;
        for(int k=1; k<=n; k++)
        {
            dis2[k]=min(dis2[k],dis2[flag]+tu[k][flag]);
        }
    }
    int ans=-1;
    for(int i=1;i<=n;i++)
    {
        ans=max(ans,dis1[i]+dis2[i]);
    }
    printf("%d\n",ans);
}